Question

The area bounded by the curves $y=\sqrt{x},2y+3=x$ and $x-$axis(in the first quadrant) is

Answer 1

The given curves are
$y=\sqrt{x}$ ................$\left(1\right)$
$2y+3=x$ ................$\left(2\right)$
and $x-$axis $y=0$ .................$\left(3\right)$
Solving eqns$\left(1\right)$ and $\left(2\right)$,we get
$2y+3=y^2$
$\Rightarrow y^2-2y-3=0$
$\Rightarrow (y-3)(y+1)=0$
$\Rightarrow y=3$ as $y\ne$negative value.
Substituting $y=3$ in $\left(1\right)$ we get
$3=\sqrt{x}$ or $x=9$
$\therefore$Required Area$={\int }_0^3(2y+3-y^2)dy$
$={[y^2+3y-\frac{y^3}{3}]}_0^3$
$=[9+9-9]=9$.sq.unit